Math Problem Statement

Rewrite the equations y^2=x(p+(px/2a))and y 2=x(p− (px/ 2a)) for the hyperbola and ellipse, respectively, in the current standard forms for those equations. What point is the center of the curve? Show in the case of the ellipse, where 2b is the length of the minor axis, that b2 = pa/2.

Solution

To rewrite the given equations for the hyperbola and ellipse in standard form, we start with the provided equations:

  1. Hyperbola: y2=x(p+px2a)y^2 = x\left(p + \frac{px}{2a}\right) We can rearrange this equation as follows: y2=px+px22ay^2 = px + \frac{px^2}{2a} Factoring out pp: y2=p(x+x22a)y^2 = p\left(x + \frac{x^2}{2a}\right) To get this in the standard form of a hyperbola, we express it as: y2p=x+x22a\frac{y^2}{p} = x + \frac{x^2}{2a}

    Rearranging gives: y2px22a=1\frac{y^2}{p} - \frac{x^2}{2a} = 1 Thus, the standard form of the hyperbola is: y2px22a=1\frac{y^2}{p} - \frac{x^2}{2a} = 1 The center of the hyperbola is at the origin (0,0)(0, 0).

  2. Ellipse: y2=x(ppx2a)y^2 = x\left(p - \frac{px}{2a}\right) Similarly, rearranging this equation gives: y2=pxpx22ay^2 = px - \frac{px^2}{2a} Factoring out pp: y2=p(xx22a)y^2 = p\left(x - \frac{x^2}{2a}\right) To write this in standard form for an ellipse, we start by dividing through by pp: y2p=xx22a\frac{y^2}{p} = x - \frac{x^2}{2a}

    Rearranging to isolate xx: y2p+x22a=x\frac{y^2}{p} + \frac{x^2}{2a} = x Reorganizing this gives us: x22a+y2p=x\frac{x^2}{2a} + \frac{y^2}{p} = x Completing the square for the xx terms yields: (xa)2a2+y2p=1\frac{(x - a)^2}{a^2} + \frac{y^2}{p} = 1 Thus, the standard form of the ellipse is: (xa)2a2+y2p=1\frac{(x - a)^2}{a^2} + \frac{y^2}{p} = 1 The center of the ellipse is at the point (a,0)(a, 0).

Showing b2=pa2b^2 = \frac{pa}{2}:

In the standard form of the ellipse:

  • The length of the semi-major axis is aa and the semi-minor axis is bb.
  • From the standard form: b2=p2b^2 = \frac{p}{2}

Now, if we set 2b2b as the length of the minor axis, it follows that b=p2b = \frac{p}{2}.

Hence, we arrive at: b2=p2=pa2b^2 = \frac{p}{2} = \frac{pa}{2}

Summary

  • Hyperbola Standard Form: y2px22a=1\frac{y^2}{p} - \frac{x^2}{2a} = 1, center (0,0)(0, 0).
  • Ellipse Standard Form: (xa)2a2+y2p=1\frac{(x - a)^2}{a^2} + \frac{y^2}{p} = 1, center (a,0)(a, 0).
  • Relationship: b2=pa2b^2 = \frac{pa}{2}.

Do you want further details or have any questions? Here are some related questions to consider:

  1. What are the properties of hyperbolas compared to ellipses?
  2. How do the foci of the ellipse and hyperbola differ?
  3. Can you explain the significance of pp in these equations?
  4. How would you derive the eccentricity of these conic sections?
  5. What other forms can these equations take based on transformations?

Tip: Always check the discriminant of the conic to determine its type!

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Math Problem Analysis

Mathematical Concepts

Conic Sections
Hyperbolas
Ellipses
Algebra

Formulas

\frac{y^2}{p} - \frac{x^2}{2a} = 1
\frac{(x - a)^2}{a^2} + \frac{y^2}{p} = 1
b^2 = \frac{pa}{2}

Theorems

Standard forms of conic sections
Properties of ellipses and hyperbolas

Suitable Grade Level

Grades 11-12